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podryga [215]
3 years ago
11

what are the benefits of VolP? select all that apply. A:cheaper printings B:clearer calls C: faster download D: increased effici

ency E: reduced cost​
Computers and Technology
2 answers:
Leya [2.2K]3 years ago
4 0

Answer:

  • Clearer calls
  • Increased efficiency
  • Reduced cost​

Explanation:

I just did this question on my online school and it was right

zzz [600]3 years ago
3 0

A has nothing to do with VoIP. B may be correct assuming your network has enough bandwidth to handle the volume of calls, and that you are able to prioritize voip traffic. d is correct; voip phones have lots of cool features like call forwarding, call queues, and voicemail. e is also correct; when using VoIP there is no charge for long distance calls, and costly pbx systems are not needed.

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He made me so angry when he didn’t let me make my point that I had to stalk after him.
madam [21]

Answer:

21

Explanation:

because

4 0
3 years ago
Read 2 more answers
Respond to the following in three to five sentences. Select the workplace skill, habit, or attitude described in this chapter (a
Katyanochek1 [597]

Answer:

communication and proactive

Explanation:

Communication is the most important thing and not only for a successful employee, is for successful life, in a company, there are a lot of people working in different areas, and sometimes we must interact with other departments to complete the task, a good communication help us to create a good experience and relationship with our partners, we're going to get efficient results.

If we want to ask more resources to our boss, we must show the needs and the benefits that company going to get, with a good presentation, inform and communication this can be easy to do, or if we are the boss, and we want to give trust our employees, we must have a good communication skills.

If we are proactive, we are going to result in almost every problem, in some companies the training is bad, we must learn process and attitudes for our job, some people are stingy with the knowledge, and we must find solution for our self, in this way we're going to be more independent, and our value increase in the company and even in the working market.

7 0
3 years ago
What does Stand for in web design
myrzilka [38]

Answer:

g

Explanation:

h

7 0
3 years ago
Create an application named Percentages whose main() method holds two double variables. Assign values to the variables. Pass bot
lbvjy [14]

Answer:

The codes below implement the problem statements

Explanation:

public class Percentages {

public static void computePercent (int a,int b)

{

System.out.println(a+" is "+(a*100/b)+"% of "+b);    

}

 

public static void main(String []args)

{

int a=2;

int b=5;

computePercent(a,b);

computePercent(b,a);

}

}

<u> </u>

<u>Part(b) </u>

import java.util.*;

public class Percentages {

public static void computePercent (int a,int b)

{

System.out.println(a+" is "+(a*100/b)+"% of "+b);

}

 

public static void main(String []args)

{

Scanner s= new Scanner(System.in);

int a=s.nextInt();

int b=s.nextInt();

computePercent(a,b);

computePercent(b,a);

}

}

8 0
3 years ago
12.2 Sort an array in ascending order Write a program that gets a list of integers from input, and outputs the integers in ascen
notsponge [240]

Answer:

Explanation:

The following code is written in Java, it asks the user for the number of inputs that will be made and then the inputs. These inputs are placed into an array. Then the array is sorted and printed out to the terminal.

public static void sortArrayList() {

           ArrayList<Integer> sortedList = new ArrayList();

           Scanner input = new Scanner(System.in);

           System.out.println("Enter number of digits in Array");

           int numberOfInputs = input.nextInt();

           for (int i = 0; i < numberOfInputs; i++) {

                   System.out.println("Enter a number");

                   sortedList.add(input.nextInt());

           }

           int n = sortedList.size();

           for (int i = 0; i < n; i++)

               for (int j = 0; j < n-i-1; j++)

                   if (sortedList.get(j) > sortedList.get(j + 1))

                   {

                       // swap arr[j+1] and arr[j]

                       int temp = sortedList.get(j);

                       sortedList.set(j, sortedList.get(j + 1));

                       sortedList.set(j + 1, temp);

                   }

           for (int x : sortedList) {

               System.out.println(x);

           }

       }

7 0
3 years ago
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